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notsponge [240]
3 years ago
5

When scientists are ready to publish the results of their experiments, why is it important for them to include a description of

the procedures they used?
Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0

Answer: To be able to not forget what have they done when they want to re-do it.

to make it easier for the next person to re-do the experiment.

Explanation:

the description of the procedure ensures that when the experiment  becomes of a success,specific steps are recorded for improvement of the experiment if maybe the experiment can be done at a short space of time.

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Since the logarithmic scale has the base 10, a change by 1 on the scale is a consequence of multiplication/division of the H+ concentration by a factor of 10. As the scale is inverse, this means that a decrease of concentration by factor 1000 is equivalent to increasing the pH by (1000/10) = 3.
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33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b
Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

Volume of the solution = V = 40.00 mL

Density =\frac{Mass}{Volume}

a) Density of the solution:

\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol

Molar mass of water = 18.02 g/mol

Moles of water= n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol

Mole fraction of fructose in this solution:\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}

\chi_1=0.1510

Mole fraction of water = \chi_2=1-\chi_1=0.8490

c) Average molar mass of of the solution:

=\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol

=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

\frac{\text{Average molar mass}}{\text{Density of the mass}}

=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol

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